Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int const N = 1e5 + 1;
struct line {
  int sum;
  line *l, *r;
  line(int val) : l(nullptr), r(nullptr), sum(val) {}
  line(line *l, line *r) : l(l), r(r), sum(0) {
    if (l) sum += l->sum;
    if (r) sum += r->sum;
  }
  line(line *v) { l = v->l, r = v->r, sum = v->sum; }
} * t[N];
line *build(int l, int r) {
  if (r == l) return new line(0);
  int m = l + r >> 1;
  return new line(build(l, m), build(m + 1, r));
}
line *update(line *v, int l, int r, int pos) {
  if (l == r) return new line(v->sum + 1);
  int m = l + r >> 1;
  line *u = new line(v);
  if (pos <= m)
    u->l = update(v->l, l, m, pos);
  else
    u->r = update(v->r, m + 1, r, pos);
  u->sum = u->l->sum + u->r->sum;
  return u;
}
int tin[N], tout[N], p[N], up[N][50], timer, n, m, q;
vector<int> g[N], all[N];
void dfs(int x, int pr) {
  tin[x] = timer++, p[x] = pr, up[x][0] = pr;
  for (int i = 1; i <= 20; i++) up[x][i] = up[up[x][i - 1]][i - 1];
  for (int to : g[x])
    if (to != pr) dfs(to, x);
  tout[x] = timer++;
}
void go(int x, int pr) {
  t[x] = t[pr];
  for (int to : all[x]) t[x] = update(t[x], 1, N, to);
  for (int to : g[x])
    if (to != pr) go(to, x);
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int l = 20; l >= 0; l--)
    if (!upper(up[a][l], b) && up[a][l]) a = up[a][l];
  return up[a][0];
}
int get(line *pr, line *lc, line *x, line *y, int l, int r, int pos) {
  if (l == r) return l;
  int m = l + r >> 1;
  int lsum = x->l->sum + y->l->sum - lc->l->sum - pr->l->sum;
  if (pos <= lsum)
    return get(pr->l, lc->l, x->l, y->l, l, m, pos);
  else
    return get(pr->r, lc->r, x->r, y->r, m + 1, r, pos - lsum);
}
int get_sz(line *x, line *y, line *lc, line *pr) {
  return x->sum + y->sum - lc->sum - pr->sum;
}
int main() {
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    cin >> x >> y, g[x].push_back(y), g[y].push_back(x);
  }
  t[0] = build(1, N);
  for (int i = 1; i <= m; i++) {
    int v;
    cin >> v, all[v].push_back(i);
  }
  dfs(1, 0), go(1, 0);
  while (q--) {
    int x, y, k;
    cin >> x >> y >> k;
    int lc = lca(x, y);
    k = min(k, get_sz(t[x], t[y], t[lc], t[p[lc]]));
    cout << k << " ";
    for (int i = 1; i <= k; i++)
      cout << get(t[p[lc]], t[lc], t[x], t[y], 1, N, i) << " ";
    cout << "\n";
  }
  return 0;
}